Two similar springs P and Q have spring constants `K_p` and `K_Q`, such that `K_p > K_Q`. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs `W_P` and `W_Q` are related as, in case (a) and case (b) respecitively

Options

A. `W_P gt W_Q , W_Q gt W_P`
B. `W_P lt W_Q , W_Q lt W_P`
C. `W_P lt W_Q , W_Q lt W_P`
D. `W_P = W_Q , W_Q = W_P`

Answer

A

Explanation

Here , `K_p > K_Q`
Case (a) : Elongation (x) in each spring is same.
`W_P = 1/2 K_px^2 , W_Q = 1/2 K_Q x^2 :. W_P > W_Q`
Case (b) : Force of elongation is same.
So, `x_1 = F/(K_P) and x_2 = F/(K_Q)`
`W_P = 1/2 K_p x_1^2 = 1/2 (F^2)/(K_P) , W_Q = 1/2 K_Q x_2^2 = 1/2 (F^2)/(K_Q)`
`:. W_P < W_Q` .